EMT 110
Engineering Materials
Lecture 2:
Atom and Bonding
Semester 1 2012/2013
Atomic Structure
Fundamental Concept
Atoms are the structural unit of all engineering
materials!
 Each atoms consist of nucleus composed of protons and
neutron and surrounded by electrons

Nucleus = Proton (positively charged particles)
+ Neutrons (no electrical charge particles)
n=1
n=2
n = Quantum number:
representing the energy level of the electron
( n, energy)
Electrons
(negatively charged particles)
Mass, Charge & Charge Unit of the
Proton, Neutron, and Electron
Particle
Proton
Neutron
Electron
Mass (g)
Charge (C)
1.67262X10-24 +1.6022X10-19
1.67493X10-24
0
9.10939X10-28 -1.6022X10-19
Charge
unit
+1
0
-1
Source: Smith, W.F. and Hashemi, J., Foundations of Materials Science and Engineering, McGraw-Hill, 2011
 Atomic
number, Z: Number of protons (p). In a
neutral atom the atomic number is equal to the
number of electrons (Z=e).
 Atomic mass, A: Total mass of proton and
neutron in the nucleus ( A=Z+N ).
 Isotope: atoms that have two or more atomic
mass. Same number of proton but different number
of neutron.
 One atomic mass unit (amu) = 1/12 of the
atomic mass of carbon
 One mole= 6.023 x 1023 atoms ( Avogadro’s
number NA ).
Z
A
Example:
Determine the number of electron (e), neutron (N)
in fluorin atom
9
19
Answer: A=p + N=19
Z=p=e= 9
So, N=A - Z=10
Atomic Number and Atomic Mass


Atomic Number = Number of Protons in the nucleus
Unique to an element


Relative atomic mass = Mass in grams of 6.203 x 1023
( Avagadro Number) Atoms.



Example :- Hydrogen = 1, Uranium = 92
Example :- Carbon has 6 Protons and 6 Neutrons. Atomic Mass =
12.
One Atomic Mass unit (amu) is 1/12th of mass of
carbon atom, indicates that the mass of one neutron
or one proton very close to one amu.
One gram mole = Gram atomic mass of an element.
Example :-
One gram
Mole of
Carbon (mol)
12 Grams
of Carbon
6.023 x 1023
Carbon
Atoms
Periodic Table
Source: Davis, M. and Davis, R., Fundamentals of Chemical Reaction Engineering, McGraw-Hill, 2003.
Example:
1) One mole aluminium have mass of 26.98 g and 6.023 x 1023 atoms.
What is the mass in grams of one atom of aluminium (A=26.98g/mol)
2) How many atom of Copper (Cu) in one gram of
Copper?(A=63.54g/mol)
Ans 1:
1 mol = 6.023 x 1023 atom
mass 1 mol Al = 26.98 g
mass (g) in 1 atom Al= 26.98 g
6.023 x 1023
Ans 2:
1 mol Cu= 63.54g
1 mol Cu= 6.023 x 1023atom
Number of Atom Cu in 1 gram Cu = 6.023 x 1023 atom
63.54
Example:
The Cladding (outside layers) of the US quarter coin consists of
an alloy of 75 wt % copper and 25 wt % nickel. What are the
atomic percent of Cu and Ni of this materials?
Solutions:
Using the basis of 100g of the 75 wt % Cu and 25 wt % Ni alloy, there are 75g
Cu and 25g Ni. Thus, the number of gram-moles of copper and nickel is
No. of gram-moles of (mol of )Cu=
75g
 1.1803mol
63.54g/mol
No. of gram moles of (mol of) Ni =
25g
 0.4260 mol
58.69g/mol
Total gram-moles= mol of Cu + mol of Ni= 1.1803 + 0.4260=1.6063 mol
Atomic % Cu=
Atomic % Ni=
1.1803mol
(100%)  73.5%
1.6063mol
0.4260mol
(100%)  26.5%
1.6063mol
Electronic Structure of Atoms




Electron rotates at definite energy levels.
Energy is absorbed to move to higher energy level.
Energy is emitted during transition to lower level.
Energy change due to transition = ΔE = hv =
Absorb
Energy
(Photon)
Emit
Energy
(Photon)
hc

h=Planks Constant
= 6.63 x 10-34 J.s
c = Speed of light
λ = Wavelength of light
v=frequency of photon
Energy levels

Energy is released or transmitted in the form of electromagnetic
radiation known as photon.
Electromagnetic Spectrum

Example
Calculate the energy in joules (J) and electron volts (eV) of the
photon whose wave length  is 121.6nm. (Given 1.00eV=1.60X10-19J;
h= 6.63X10-34J.s)
Answer:
ΔE =
hc

(6.63 10 34 J.s)(3.00 108 m/s )
E 
(121.6nm)(10-9 m/nm)
 1.63 10 18 J
 1.63 10-18 J 
1eV
 10.2eV
-19
1.60 10 J
Energy in Hydrogen Atom


Hydrogen atom has one proton and one electron
Energy of hydrogen atoms for different energy levels is given
by
(n=1,2…..) principal quantum
13.6
E   2 eV
numbers
n

Example:- If an electron undergoes transition from n=3 state
to n=2 state, the energy of photon emitted is
E  

13.6
2
3
 (
13.6
2
2
)  1.89ev
Energy required to completely remove an electron from
hydrogen atom is known as ionization energy
Energy level diagram for hydrogen
This diagram explains the energy level changes
for a hydrogen atom, and which transitions
cause which type of light. The horizontal lines
on the left represent the different energy
levels. The red arrows represent the drops an
electron can take.
Let’s say an excited electron is in n=3. From
there, that electron can drop down to n=2 or
n=1. The transition of n=3 to n=2 produces a
red band of light. The transition of n=3 to n=1
produces UV light. These two types of light,
along with the others can be detected when
electricity is run through a sample of hydrogen
gas.

A hydrogen atom exists with its electron in the n= 3 state. The
electron undergoes a transition to the n=2 state. Calculate (a)
the energy of the photon emitted, (b) its frequency, and (c) its
wavelength, (d) energy is absorbed or emitted, and (e) which
series it belong to and what type emission does it represent?
Ans: (a) Energy of the photon emited is:
 13.6eV
E
n2
E  E3  E2
- 13.6  13.6
 2
2
3
2
 1.89eV

1.60 10-19 J
 1.89eV 
eV
 3.02 10-19 J
b) The frequency of the photon is
E  hv
E
v
h
3.02 10-19 J

6.63 10 34 J.s
 4.55 1014 Hz
c) The wavelength of the photon is
hc (6.63 10 34 J.s)(3.00 108 m/s )


E
3.02 10 19 J
 6.59 10-7 m
1nm
 6.59 10 m  -9
10 m
 659nm
-7
d) Energy is released as its quantity is positive, and the electron
is transitioning from a higher orbit to lower orbit.
c) The emission belongs to Balmer series and corresponds to
visible red light.
Balmer series
Visible red light
(690 nm)
Quantum Numbers of Electrons of Atoms
Subsidiary Quantum
Number (l)
Principal Quantum Number
(n)



Represents main energy
levels.
Range of n from 1 to 7.
Larger the ‘n’ higher the
energy.



Represents sub energy
levels (orbital).
Range of l from 0 to n-1.
Represented by letters
s,p,d and f.
n=1
n=2
n=3
n=2
s orbital
(l=0)
n=1
p Orbital
(l=1)
Quantum Numbers of Electrons of
Atoms (Cont..)




2-9
Magnetic Quantum
Number ml
Represents spatial
orientation of single
atomic orbital.
Permissible values are –l
to +l.
Example:- if l=1,
ml = -1,0,+1.
I.e. 2l+1 allowed values.
No effect on energy.





Electron spin quantum
number ms
Specifies two directions
of electron spin.
Directions are clockwise
or anticlockwise.
Values are +1/2 or –1/2.
Two electrons on same
orbital have opposite
spins.
No effect on energy.
Relationship between principal (n),
subsidiary (l) and magnetic (ml)
quantum numbers
Example:
l = 0 to (n -1), e.g: if n=3, then l= 0 to (3-1=2) or l=(0, 1, 2)
If l=0-->s orbital, l=1-->p orbital, l=2-->d orbital, l=3-->f orbital
ml = -l to +l, if l = 2, so ml = -2 to +2 or ml = (-2, -1, 0, +1, +2)
Electron Structure of Multielectron Atom



Maximum number of electrons in each atomic shell
is given by 2n2.
Atomic size (radius) increases with addition of shells.
Electron Configuration lists the arrangement of
electrons in orbitals.
Number of Electrons

Example :-
Orbital letters
1s2 2s2 2p6 3s2
Principal Quantum Numbers

2-10
For Iron, (Z=26), Electronic configuration is
1s2 2s2 2p6 3s2 3p6 3d6 4s2
Stable electron configuration
Stable electron configurations...
 have complete s and p subshells
 tend to be unreactive.
Most elements: Electron configuration not stable!
Electron configuration
1s1
1s2
(stable)
1s22s1
1s22s2
1s22s22p1
1s22s22p2
...
1s22s22p6
(stable)
1s22s22p63s1
1s22s22p63s2
1s22s22p63s23p1
...
1s22s22p63s23p6
(stable)
...
1s22s22p63s23p63d10 4s246 (stable)
• Why? Valence (outer) shell usually not filled completely.
Example

Write the electron configuration for the following atoms by
using conventional spdf notation.

a) Fe atom (Z=26) and the Fe2+ and Fe3+ ions
remember !!!!!
7s 7p 7d
6s 6p 6d 6f
5s 5p 5d 5f
4s 4p 4d 4f
3s 3p 3d
2s 2p
1s
Electrons fill up in this order
The order for writing the orbitals
Electron Structure and Chemical
Activity (Cont..)



Electronegative elements accept electrons during
chemical reaction.
Some elements behave as both electronegative and
electropositive.
Electronegativity is the degree to which the atom
attracts electrons to itself


Measured on a scale of 0 to 4.1
Example :- Electronegativity of Fluorine is 4.1
Electronegativity of Sodium is 1.
Te
Na
Electropositive 0
K 1
W
2H
N
Se
3
O
Fl
4
Electronegative
Atomic and Molecular Bonds





2-12
Ionic bonds :- Strong atomic bonds due to transfer of
electrons
Covalent bonds :- Large interactive force due to
sharing of electrons
Metallic bonds :- Non-directional bonds formed by
sharing of electrons
Permanent Dipole bonds :- Weak intermolecular
bonds due to attraction between the ends of
permanent dipoles.
Fluctuating Dipole bonds :- Very weak electric dipole
bonds due to asymmetric distribution of electron
densities.
Ionic Bonding



Ionic bonding is due to electrostatic (Coulombic)
force of attraction between cations and anions.
It can form between metallic and nonmetallic
elements.
Electrons are transferred from electropositive to
electronegative atoms
Electropositive
Electronegative
Electron
Element
Atom
Transfer
Cation
+ve charge
Electrostatic
Attraction
IONIC BOND
2-14
Anion
-ve charge
Ionic Bonding



Lattice energies and melting points of ionically bonded solids are
high.
Lattice energy decreases when size of ion increases.
Multiple bonding electrons increase lattice energy.
Chlorine
Atom
Cl
Sodium
Atom
Na
Sodium Ion
Na+
I
O
N
I
C
B
O
N
D
Chlorine Ion
Cl -
Ionic Force for Ion Pair


Nucleus of one ion attracts electron of another ion.
The electron clouds of ion repulse each other when
they are sufficiently close.
Force versus separation
Distance for a pair of
oppositely charged ions
Figure 2.11
2-16
Ion Force for Ion Pair (Cont..)
 Z e Z e   Z Z e


F
4  a  4  a 
2
1
2
2
attractive
Z1,Z2 = Number of electrons removed or0
added during ion formation
e = Electron Charge
a = Interionic seperation distance
ε = Permeability of free space (8.85 x 10-12c2/Nm2)
1
2
2
0
(n and b are constants)
F
repulsive

nb
a
n 1
2
Fnet  Fattractive  Frepulsive
2-17
F net  
Z Z e  nb
4  a  a
1
2
2
0
n 1
Interionic Force - Example

Force of attraction between Na+ and Cl- ions
Z1 = +1 for Na+, Z2 = -1 for Cle = 1.60 x 10-19 C , ε0 = 8.85 x 10-12 C2/Nm2
a0 = Sum of Radii of Na+ and Cl- ions
= 0.095 nm + 0.181 nm = 2.76 x 10-10 m
Cl-
Na+
a0
19
( 1)( 1)(1.60  10 C )
e
Z
Z


4  a  4 (8.85 x 10 C /Nm2)(2.76 x 10
2
F
1
attraction
 3.02  10 9 N
2
2
0
2-18
2
-12
2
-10
m)
Interionic Energies for Ion Pairs

Net potential energy for a pair of oppositely
charged ions =
2
b
e
Z
Z

E 
4  a  a
1
2
net
2
n
0
Attraction Repulsion
Energy Energy
Energy
Energy
Released Absorbed

2-19
Enet is minimum when ions are at equilibrium
seperation distance a0
Ion Arrangements in Ionic Solids


Ionic bonds are Non Directional
Geometric arrangements are present in solids to
maintain electric neutrality.

Example:- in NaCl, six Cl- ions pack around central Na+ Ions
Ionic packing
In NaCl
and CsCl
Figure 2.13
CsCl

2-20
NaCl
As the ratio of cation to anion radius decreases,
fewer anion surround central cation.
Bonding Energies



Lattice energies and melting points of ionically
bonded solids are high.
Lattice energy decreases when size of ion
increases.
Multiple bonding electrons increase lattice
energy.

Example :NaCl
CsCl
BaO
2-21
Lattice energy = 766 KJ/mol
Melting point = 801oC
Lattice energy = 649 KJ/mol
Melting Point = 646oC
Lattice energy = 3127 KJ/mol
Melting point = 1923oC
Covalent Bonding



In Covalent bonding, outer s and p electrons are shared
between two atoms to obtain noble gas configuration.
Takes place between elements with small differences in
electronegativity and close by in periodic table.
In Hydrogen, a bond is formed between 2 atoms by
sharing their 1s1 electrons
Electron
Pair
H
+ H
1s1
Electrons
H H
Hydrogen
Molecule
Overlapping Electron Clouds
Covalent Bonding - Examples


In case of F2, O2 and N2, covalent bonding is formed by
sharing p electrons
Fluorine gas (Outer orbital – 2s2 2p5) share one p electron to
attain noble gas configuration.
F

+ F
H
F F
F F
Bond Energy=160KJ/mol
Oxygen (Outer orbital - 2s2 2p4) atoms share two p electrons
O + O
O
O
O=O
Bond Energy=28KJ/mol

Nitrogen (Outer orbital - 2s2 2p3) atoms share three p electrons
N + N
2-23 H H
N
N
N
N
Bond Energy=54KJ/mol
Covalent Bonding in Benzene



Chemical composition of Benzene is C6H6.
The Carbon atoms are arranged in hexagonal ring.
Single and double bonds alternate between the atoms
H
C
H
C
H
H
C
C
C
H
H
Structure of Benzene
Simplified Notations
Metallic Bonding





Atoms in metals are closely packed
in crystal structure.
Loosely bounded valence electrons
are attracted towards nucleus of
other atoms.
Electrons spread out among atoms
forming electron clouds.
These free electrons are reason for
electric conductivity and thermal
conductivity
Since outer electrons are shared by
many atoms,
metallic bonds are Non-directional
Positive
Ion
Valence electron charge cloud
Metallic Bonds (Cont..)



Overall energy of individual atoms are lowered by
metallic bonds
Minimum energy between atoms exist at equilibrium
distance a0
Fewer the number of valence electrons involved, more
metallic the bond is.

Example:- Na
Bonding energy 108KJ/mol,
Melting temperature 97.7oC

Higher the number of valence electrons involved,
higher is the bonding energy.

Example:-
Ca
Bonding energy 177KJ/mol,
Melting temperature 851oC
2-29
Secondary Atomic Bonding




Secondary bonds are due to attractions of electric dipoles in atoms or
molecules.
Dipoles are created when positive and negative charge centers exist.
Bonding result from the columbic attraction between positive end of
one dipole and the negative region of an adjacent one
Sometimes called Van der Waals bond
Dipole moment=μ =q.d
+q
q= Electric charge
d = separation distance
d

There two types of bonds
permanent and fluctuating.
Fluctuating Dipole Bond



Weak secondary bonds in noble gasses.
Dipoles are created due to asymmetrical distribution of
electron charges.
Electron cloud charge changes with time.
Symmetrical
distribution
of electron charge
Asymmetrical
Distribution
(Changes with time)
Permanent Dipole Bond

Secondary bond created by the attraction of molecules that
have permanent dipole. That is, each molecule has positive and
negative charge center separated by distance.
CH4
CH3Cl
Symmetrical
Arrangement
Of 4 C-H bonds
No Dipole
moment
Asymmetrical
Tetrahedral
arrangement
Creates
Dipole
Hydrogen Bond


Hydrogen bonds are Dipole-Dipole interaction between polar
bonds containing hydrogen atom.
Special type of intermolecular permanent dipole attraction
that occur between hydrogen atom bonded to a highly
electronegative element (F, O, N or Cl) and another atom of a
highly electronegative element.
H
105 0
O
H
Hydrogen
Bond
Mixed Bonding

Chemical bonding of atoms or ions can involve more than
one type of primary bond and can also involve secondary
dipole bond.
1.
Ionic - covalent
Metallic – covalent
Metallic – ionic
Ionic – covalent - metallic
2.
3.
4.